If (1 -x + 2x^2)^n = a_0 + a_1x + a_2x^2+.... | vedclass

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Question

$\left(1-\mathrm{x}+2 \mathrm{x}^{2}\right)^{\mathrm{n}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots$    &

Vedclass · Accepted Answer

no value of $n$

Vedclass · Answer

$\left(1-\mathrm{x}+2 \mathrm{x}^{2}ight)^{\mathrm{n}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots$          .....$(1)$

$\mathrm{x}=0 \Rightarrow \mathrm{a}_{0}=1$

Diff. $(1)$

$n\left(1-x+2 x^{2}ight)^{n-1}(-1+4 x)$

$=a_{1}+2 a_{2} x+\ldots \ldots \ldots$          .........$(2)$

put $\mathrm{x}=0$

$-\mathrm{n}=\mathrm{a}_{1}$

Diff. $(2)$

${\mathrm{n}(\mathrm{n}-1)\left(1-\mathrm{x}+2 \mathrm{x}^{2}ight)^{\mathrm{n}-2}(-1+4 \mathrm{x})^{2}} $

${+\mathrm{n}\left(1-\mathrm{x}+2 \mathrm{x}^{2}ight)^{\mathrm{n}-1}(4)} $

${\quad=2 \mathrm{a}_{2}+\ldots \ldots \ldots}$         .........$(3)$

Put $\mathrm{x}=0$

$\mathrm{n}^{2}-\mathrm{n}+4 \mathrm{n}=2 \mathrm{a}_{2}$

$2 \mathrm{a}_{2}=\mathrm{n}^{2}+3 \mathrm{n}$

$\mathrm{a}_{1}+\mathrm{a}_{0}=2 \mathrm{a}_{2}$

$\Rightarrow-\mathrm{n}+1=\mathrm{n}^{2}+3 \mathrm{n}$

$\mathrm{n}^{2}+4 \mathrm{n}-1=0$

$\Rightarrow \mathrm{n} 
otin \mathrm{N}$

If $(1 -x + 2x^2)^n$ = $a_0 + a_1x + a_2x^2+..... a_{2n}x^{2n}$ , $n \in N$ , $x \in R$ and $a_0$ , $a_2$ and $a_1$ are in $A$ . $P$ .,then there exists

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